CODEWITHSUNDEEP
pythonscrapy
Shiva Krishna Bavandla
Shiva Krishna Bavandla

Reputation: 26698

Constructing a regular expression for url in start_urls list in scrapy framework python

I am very new to scrapy and also i didn't used regular expressions before

The following is my spider.py code

class ExampleSpider(BaseSpider):
   name = "test_code
   allowed_domains = ["www.example.com"]
   start_urls = [
       "http://www.example.com/bookstore/new/1?filter=bookstore",
       "http://www.example.com/bookstore/new/2?filter=bookstore",
       "http://www.example.com/bookstore/new/3?filter=bookstore",
   ]

   def parse(self, response):
       hxs = HtmlXPathSelector(response)

Now if we look at start_urls all the three urls are same except they differ at integer value 2?, 3? and so on i mean unlimited according to urls present on the site , i now that we can use crawlspider and we can construct regular expression for the URL like below, 

    from scrapy.contrib.spiders import CrawlSpider, Rule
    from scrapy.contrib.linkextractors.sgml import SgmlLinkExtractor
    import re

    class ExampleSpider(CrawlSpider):
        name = 'example.com'
        allowed_domains = ['example.com']
        start_urls = [
       "http://www.example.com/bookstore/new/1?filter=bookstore",
       "http://www.example.com/bookstore/new/2?filter=bookstore",
       "http://www.example.com/bookstore/new/3?filter=bookstore",
   ]

        rules = (
            Rule(SgmlLinkExtractor(allow=(........),))),
        ) 

   def parse(self, response):
       hxs = HtmlXPathSelector(response)

can u please guide me , that how can i construct a crawl spider Rule for the above start_url list.

Upvotes: 1

Views: 4430

Answers (2)

warvariuc
warvariuc

Reputation: 59644

If i understand you correctly, you want a lot of start URL with a certain pattern.

If so, you can override BaseSpider.start_requests method:

class ExampleSpider(BaseSpider):
    name = "test_code"
    allowed_domains = ["www.example.com"]

    def start_requests(self):
        for i in xrange(1000):
            yield self.make_requests_from_url("http://www.example.com/bookstore/new/%d?filter=bookstore" % i)

    ...

Upvotes: 4

pjob
pjob

Reputation: 90

If you are using CrawlSpider, it's not usually a good idea to override the parse method.

Rule object can filter the urls you are interesed to the ones you do not care for.

See CrawlSpider in the docs for reference. 

from scrapy.contrib.spiders import CrawlSpider, Rule
from scrapy.contrib.linkextractors.sgml import SgmlLinkExtractor
import re

class ExampleSpider(CrawlSpider):
    name = 'example.com'
    allowed_domains = ['example.com']
    start_urls = ['http://www.example.com/bookstore']

    rules = (
        Rule(SgmlLinkExtractor(allow=('\/new\/[0-9]\?',)), callback='parse_bookstore'),
    )

def parse_boostore(self, response):
   hxs = HtmlXPathSelector(response)

Upvotes: 0

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